A 5.00 L sample of gas at 1.00 atm and 20 degrees Celsius is compressed to 2.00 L at 2.00 atm. What is the temperature of the gas?

234.4 K

Explanation:

Using the combined gas law equation as follows;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (atm)

P2 = final pressure (atm)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the provided information,

P1 = 1.00atm

P2 = 2.00 atm

V1 = 5.00L

V2 = 2.00L

T1 = 20°C = 20 + 273 = 293K

T2 = ?

Using P1V1/T1 = P2V2/T2

1 × 5/293 = 2 × 2/T2

5/293 = 4/T2

Cross multiply

293 × 4 = 5 × T2

1172 = 5 T2

T2 = 1172 ÷ 5

T2 = 234.4 K

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.